Browse Trigonometry

$OAB$ is a sector of a circle with centre $O$ and radius $r$ cm
angle $AOB = 0.75$ radians
the arc $AB$ has length $10.5$ cm
$OBC$ is a triangle joined to the sector along the radius $OB$ , where $BC = 9$ cm and angle $OBC = 2.1$ radians

(a) Show that

r = 14

.

(2 marks)
(b) Find the area of the plan view of the screen, giving your answer to 3 significant figures.

(4 marks)

●●●●●Level 46 marksStart

A jeweller is designing a silver brooch.

The brooch is modelled as the region bounded by the chord

AB

and the arc

AB

of a circle with centre

O

and radius

16

cm, where angle

AOB = \dfrac{2\pi}{3}

radians.
(a) Use the cosine rule to show that

AB = 16\sqrt{3}

cm.

(2 marks)
(b) Find the exact perimeter of the brooch.

(2 marks)
(c) Find the exact area of the brooch, giving your answer in the form

(a\pi + b\sqrt{3})

^2

, where

a

and

b

are rational numbers.

(3 marks)

●●●●●Level 47 marksStart

The diagram shows a sector

OAB

of a circle with centre

O

and radius

5

cm. The angle

AOB

\theta

radians. The shaded segment is bounded by the chord

AB

and the arc

AB

, and the perimeter of the shaded segment is

21

cm.(a) Show that

\theta = 4.2 - 2\sin\left(\tfrac12\theta\right)

.

(4 marks)

The equation

\theta = 4.2 - 2\sin\left(\tfrac12\theta\right)

has only one root for

0 < \theta < \pi

.
(b) This root can be found by using the iterative formula

\theta_{n+1} = 4.2 - 2\sin\left(\tfrac12\theta_n\right)

with a starting value of

\theta_1 = 2

Write down the values of $\theta_2$ , $\theta_3$ and $\theta_4$ .

Hence find the value of this root correct to 3 significant figures.

(3 marks)

The diagram shows the graph of

y = 4.2 - 2\sin\left(\tfrac12\theta\right)

and the line

y = \theta

, for

0 \le \theta \le \pi

.(c) • Use this diagram to show how the iterative process used in (b) converges to this root.

State the type of convergence.

(3 marks)

●●●●●Level 410 marksStart

Figure 1 (not accurately drawn)

Two horizontal ropes are tied to a mooring post at the same point

O

. The ropes exert two horizontal forces,

\mathbf{P}

and

\mathbf{Q}

, on the post, and the forces are modelled as acting at

O

.

The angle between the direction of

\mathbf{P}

and the direction of

\mathbf{Q}

120°

, as shown in Figure 1.

Force $\mathbf{P}$ has magnitude $X$ newtons
Force $\mathbf{Q}$ has magnitude $60$ N
The resultant of $\mathbf{P}$ and $\mathbf{Q}$ has magnitude $20\sqrt{19}$ N

Find

(i) the value of

X

(ii) the angle between

\mathbf{Q}

and the resultant of

\mathbf{P}

and

\mathbf{Q}

, giving your answer to the nearest degree.

(8 marks)

●●●●●Level 48 marksStart

The point

A

is such that

|\overrightarrow{OA}| = 10

and

\overrightarrow{OA}

makes an angle of

135°

with the positive

x

-axis, measured anticlockwise.
(a) (i) On a sketch of the

x

- and

y

-axes, mark the position of the point

A

.

(ii) Find the position vector of

A

in terms of

\mathbf{i}

and

\mathbf{j}

.

(4 marks)

The point

B

has position vector

8\mathbf{i}

.
(b) Find the exact area of triangle

AOB

.

(2 marks)

OABC

is a parallelogram.
(c) Find the position vector of

C

in terms of

\mathbf{i}

and

\mathbf{j}

.

(2 marks)

●●●●●Level 48 marksStart

The diagram shows triangle

ABC

in which angle

BAC

120°

and the lengths of

AB

and

AC

are

(5 + h)

cm and

(3 - h)

cm respectively, where

0 < h < 3

.(a) Show that the length of

BC

\sqrt{p}

cm where

p = 49 + 2h + h^2

.

(2 marks)
(b) Hence show that, when

h

is small,

\sqrt{p} \approx 7 + mh + nh^2

, where

m

and

n

are rational numbers whose values are to be determined.

(4 marks)

●●●●●Level 46 marksStart

Figure 1

Diagram NOT accurately drawn

Figure 1 shows the plan view of a flower bed

ACD

.

The flower bed consists of triangle

BCD

joined to triangle

ABD

.

Given that

$BD = x$ m
$CD = (x + 3)$ m
$BC = 7$ m
angle $BCD = \theta°$

(a) show that

\cos \theta° = \frac{3x + 29}{7(x + 3)}

(2 marks)

Given also that

$x = 5$
angle $BAD = 35°$
$C$ , $B$ and $A$ lie on a straight line

(b) find the area of the flower bed

ACD

, giving your answer, in m², to one decimal place.

(5 marks)

●●●●●Level 47 marksStart

Figure 1

Diagram NOT accurately drawn

The shaded area in Figure 1 shows the plan view of a café terrace.

The terrace consists of a sector

AOB

of a circle centre

O

joined to a triangle

OBC

.

Given that

$OA = OB = 5$ m
$OC = 8$ m
$BC = 6$ m
$AOC$ is a straight line

(a) show that angle

BOC

0.847

radians to 3 decimal places.

(3 marks)
(b) Find the total area of the terrace. Give your answer in m

^2

to 3 significant figures.

(3 marks)
(c) Find the perimeter of the terrace. Give your answer in metres to 3 significant figures.

(3 marks)

●●●●●Level 49 marksStart

Exact values of trigonometric ratios

24 questions

LessonNot started

In this question you must show detailed reasoning.

Solve the equation

\sin\left(2x - \dfrac{\pi}{6}\right) = \cos 2x

for

0 \le x < 2\pi

, giving your answers in exact form. (5 marks)

●●●●●Level 45 marksStart

The curve

C

has parametric equations

x = 4\cos t, \qquad y = 3\sin t, \qquad 0 \leqslant t < 2\pi

(a) Find

\dfrac{\mathrm{d}y}{\mathrm{d}x}

in terms of

t

. (3 marks)
(b) The point

P

C

has parameter

t = \dfrac{\pi}{3}

.

Find an equation of the normal to

C

P

, giving your answer in the form

y = mx + c

, where

m

and

c

are exact constants. (6 marks)

●●●●●Level 49 marksStart

Solve the equation

2\cos\left(2x - \dfrac{\pi}{3}\right) = 1

for

0 \leqslant x \leqslant \pi

, giving your answers in terms of

\pi

. (5 marks)

●●●●●Level 45 marksStart

It is given that the angle

\theta

satisfies the equation

\sin(2\theta + 45^\circ) = 4\sin(2\theta - 45^\circ)

.
(a) Show that

\tan 2\theta = \dfrac53

.

(3 marks)
(b) Hence find, in surd form, the exact value of

\tan\theta

, given that

\theta

is an obtuse angle.

(5 marks)

●●●●●Level 48 marksStart

(a) Use the result

\cos(A + B) = \cos A\cos B - \sin A\sin B

to show that

\cos(A - B) = \cos A\cos B + \sin A\sin B

.

(2 marks)

The function

\mathrm{f}(\theta)

is defined as

\mathrm{f}(\theta) = 5 + 8\cos(\theta + 60)^\circ\cos(\theta - 60)^\circ

, where

\theta

is in degrees.
(b) Show that

\mathrm{f}(\theta) = 7 - 8\sin^2\theta

.

(3 marks)
(c) (i) Determine the following.

The maximum value of $\mathrm{f}(\theta)$

The smallest positive value of $\theta$ for which this maximum value occurs

(ii) Determine the following.

The minimum value of $\mathrm{f}(\theta)$

The smallest positive value of $\theta$ for which this minimum value occurs

(4 marks)

●●●●●Level 49 marksStart

It is given that

(5\sin\theta + 4\cos\theta)^2 + (2\sin\theta - 10\cos\theta)^2 = 58

and that

\theta

is obtuse.

Find the exact value of

\sin\theta

Fully justify your answer.

(6 marks)

●●●●●Level 46 marksStart

The curve with equation

y = \sin x°

for

-360 \le x \le 360

is shown below.(a) Point

A

on the curve has coordinates

\left(a, \dfrac{\sqrt{3}}{2}\right)

(a) (i) Find the value of

a

(ii) State the value of

\sin(180° - a°)

(3 marks)
(b) Point

B

on the curve has coordinates

\left(b, -\dfrac{\sqrt{7}}{4}\right)

(b) (i) Find the exact value of

\sin(b° - 180°)

(ii) Find the exact value of

\cos b°

(5 marks)

●●●●●Level 48 marksStart

A sequence

u_1, u_2, u_3, \ldots

is defined by

u_1 = 9

u_{n+1} = u_n + 3\cos\left(\frac{n\pi}{2}\right)

(a) Find the value of

u_2

, the value of

u_3

, the value of

u_4

and the value of

u_5

(2 marks)
(b) State the order of this periodic sequence.

(1 mark)
(c) Find the value of

\displaystyle\sum_{n=1}^{62} u_n

(3 marks)

●●●●●Level 46 marksStart

A jeweller is designing a silver brooch.

The brooch is modelled as the region bounded by the chord

AB

and the arc

AB

of a circle with centre

O

and radius

16

cm, where angle

AOB = \dfrac{2\pi}{3}

radians.
(a) Use the cosine rule to show that

AB = 16\sqrt{3}

cm.

(2 marks)
(b) Find the exact perimeter of the brooch.

(2 marks)
(c) Find the exact area of the brooch, giving your answer in the form

(a\pi + b\sqrt{3})

^2

, where

a

and

b

are rational numbers.

(3 marks)

●●●●●Level 47 marksStart

The first, second and fourth terms of an arithmetic progression are

u_1

u_2

and

u_4

respectively, where

u_1 = 3\sin\theta, \qquad u_2 = 5\cos\theta, \qquad u_4 = -21\sin\theta

and

\tfrac12\pi < \theta < \pi

.
(a) Determine the exact value of

\theta

.

(3 marks)
(b) Hence determine the value of

\displaystyle\sum_{r=1}^{20} u_r

.

(3 marks)

●●●●●Level 46 marksStart

Graphs of sine, cosine and tangent

31 questions

LessonNot started

(a) Use the result

\cos(A + B) = \cos A\cos B - \sin A\sin B

to show that

\cos(A - B) = \cos A\cos B + \sin A\sin B

.

(2 marks)

The function

\mathrm{f}(\theta)

is defined as

\mathrm{f}(\theta) = 5 + 8\cos(\theta + 60)^\circ\cos(\theta - 60)^\circ

, where

\theta

is in degrees.
(b) Show that

\mathrm{f}(\theta) = 7 - 8\sin^2\theta

.

(3 marks)
(c) (i) Determine the following.

The maximum value of $\mathrm{f}(\theta)$

The smallest positive value of $\theta$ for which this maximum value occurs

(ii) Determine the following.

The minimum value of $\mathrm{f}(\theta)$

The smallest positive value of $\theta$ for which this minimum value occurs

(4 marks)

●●●●●Level 49 marksStart

The depth of water,

d

metres, over a sandbar at the mouth of an estuary on one day is modelled by the equation

d = 4.5 + 2.3\cos\big(29(t - 3.2)\big)^\circ,

where

t

is the number of hours after midnight.
(a) (i) Find the least depth of water over the sandbar given by this model.

(ii) Find the value of

t

when the least depth first occurs.

(3 marks)
(b) A yacht can only cross the sandbar when the depth of water is at least

3.1

metres.

Determine the two periods of time during the day between which this yacht will not be able to cross the sandbar. Give your answers correct to the nearest minute.

(5 marks)

After heavy rain, more water flows down the estuary. An improved model uses the equation

d = 4.5 + 0.03r + 2.3\cos\big(29(t - 3.2)\big)^\circ,

where

r

is the rainfall, in millimetres, over the previous 24 hours.
(c) Explain how this new equation could give an improved model.

(1 mark)

●●●●●Level 49 marksStart

The function f is defined by

\text{f}(x) = \arccos\left(\frac{x}{2}\right) \quad \text{for } 0 \le x \le a

The curve with equation

y = \text{f}(x)

is shown below.(a) State the value of

a

(1 mark)
(b) (i) On the diagram above, sketch the curve with equation

y = 2\cos x \quad \text{for } 0 \le x \le \frac{\pi}{2}

and sketch the line with equation

y = x \quad \text{for } 0 \le x \le \frac{\pi}{2}

(ii) Explain why the solution to the equation

x - 2\cos x = 0

must also be a solution to the equation

2\cos x = \arccos\left(\frac{x}{2}\right)

(5 marks)
(c) Use the Newton-Raphson method with

x_0 = 0.5

to find an approximate solution,

x_3

, to the equation

x - 2\cos x = 0

Give your answer to four decimal places.

(3 marks)

●●●●●Level 49 marksStart

The curve with equation

y = \sin x°

for

-360 \le x \le 360

is shown below.(a) Point

A

on the curve has coordinates

\left(a, \dfrac{\sqrt{3}}{2}\right)

(a) (i) Find the value of

a

(ii) State the value of

\sin(180° - a°)

(3 marks)
(b) Point

B

on the curve has coordinates

\left(b, -\dfrac{\sqrt{7}}{4}\right)

(b) (i) Find the exact value of

\sin(b° - 180°)

(ii) Find the exact value of

\cos b°

(5 marks)

●●●●●Level 48 marksStart

(a) Given that

5\cos x + 3\sin x \equiv R\cos(x - \alpha)

, where

R > 0

and

0 < \alpha < \dfrac{\pi}{2}

, find

R

exactly and find

\alpha

, in radians, correct to 3 decimal places.

(3 marks)

An arithmetic sequence has first term

5\cos x

and common difference

\sin x

.

The sum of the first

7

terms of this sequence is

S_7

.
(b) (i) Using your answer to part (a), find the exact maximum value of

S_7

x

varies.

(ii) Write down the smallest positive value of

x

for which

S_7

takes this maximum value.

(3 marks)

●●●●●Level 46 marksStart

A wind turbine has three blades that rotate at a constant rate about a horizontal axis.

The height above the ground,

H

m, of the tip of one of the blades,

t

seconds after the turbine begins to rotate, is modelled by

H = \left|A\sin(bt + \alpha)^\circ\right|

where

A

b

and

\alpha

are constants.

In the model

the greatest height of the tip of the blade above the ground is $66$ m
the tip of the blade is $12$ m above the ground when the turbine begins to rotate
one complete rotation of the blades takes $6$ seconds, during which the tip of the blade is at its greatest height once

(a) Find a complete equation for the model, giving

A

and

b

exactly and giving

\alpha

to 3 significant figures.

(4 marks)

A student suggests that the model

H = \left|A\sin(bt + \alpha)^\circ\right| + d

where

d

is a positive constant, would be more appropriate.
(b) Explain why the student is correct.

(1 mark)

●●●●●Level 45 marksStart

The number of hours of daylight,

D

, in a town on the day that is

t

days after 21 March is modelled by the equation

D = A + B\sin(kt)^\circ

where

A

B

and

k

are positive constants.

Given that, according to the model,

$D$ is periodic with period $365$ days
the greatest number of hours of daylight in the town is $16.4$
the least number of hours of daylight in the town is $7.6$

(a) find a complete equation for the model.

(3 marks)
(b) Write down the value of

t

on which, according to the model, the town has the greatest number of hours of daylight.

(1 mark)
(c) Use the model to find the number of hours of daylight in the town when

t = 41

, giving your answer in hours and minutes to the nearest minute.

(2 marks)

On 21 December, when

t = 275

, the town actually had

7.9

hours of daylight.
(d) Use this information to comment on the model.

(2 marks)

●●●●●Level 48 marksStart

In a warehouse, storage crates are carried around a vertical loop of track on a carousel.

Each crate makes repeated circuits of the loop. In a model

the greatest height of a crate above the floor is $14$ m
each circuit begins with the crate $4$ m above the floor
the floor is horizontal

The height,

H

m, of a crate above the floor,

t

seconds after it begins its first circuit, is modelled by

H = a - b(t - 10)^2

where

a

and

b

are positive constants.
(a) Find a complete equation for the model.

(3 marks)
(b) Find, according to the model, the height of the crate above the floor when

t = 20

(1 mark)

An alternative model for the height of the crate,

t

seconds after it begins its first circuit, is

H = 5\cos(18t + \alpha)^\circ + \beta

where

\alpha

and

\beta

are constants and

0 \leqslant \alpha < 360

(c) Find a complete equation for the alternative model.

(2 marks)

The carousel runs continuously for

3

minutes.
(d) Explain why the alternative model is more suitable than the first model.

(1 mark)

●●●●●Level 47 marksStart

(a) Express

8\cos\theta + 15\sin\theta

in the form

R\cos(\theta - \alpha)

, where

R > 0

and

0 < \alpha < 90^\circ

State the value of

R

and give the value of

\alpha

, in degrees, to 2 decimal places.

(3 marks)

The air temperature,

T\ ^\circ

C, at a weather station in a desert,

t

hours after midnight on a particular day, is modelled by the equation

T = 15 - 8\cos(15t)^\circ - 15\sin(15t)^\circ \qquad 0 \leqslant t < 24

(b) Using your answer to part (a), find

(i) the maximum temperature at the weather station on that day according to the model,

(ii) the time of day at which this maximum temperature occurs according to the model, giving your answer to the nearest minute.

(3 marks)

On that day, the lowest temperature recorded at the weather station was

4\ ^\circ

C, at 06:30
(c) Use this information to comment on the model.

(2 marks)

●●●●●Level 48 marksStart

A biologist models the number,

N

, of water fleas in a pond

t

days after she starts observing them by

N = 5200 - 1800\cos(15t)^\circ.

(a) State the greatest number of water fleas in the pond according to the model.

(1 mark)
(b) Write down the difference between the greatest and least values of

N

.

(1 mark)
(c) Without giving any numerical values, describe briefly how the number of water fleas changes as

t

increases.

(1 mark)
(d) Find how long it takes for

N

to first return to its value at

t = 0

.

(2 marks)
(e) Determine the time on the second occasion when

N = 6000

.

(4 marks)

The biologist observes the pond for several months. She notices that, although the population rises and falls in a way similar to the model, each peak is larger than the peak before it.
(f) Suggest a change to the model that will take account of this observation.

(1 mark)

●●●●●Level 410 marksStart

Figure 1

Figure 1 shows a sketch of part of the curve

C_1

with equation

y = -3\sin x

where

x

is measured in radians.

The point

P

is the maximum point on

C_1

with the smallest positive

x

coordinate. The point

Q

is a minimum point on

C_1

, as shown in Figure 1.
(a) State

(i) the coordinates of

P

(ii) the coordinates of

Q

(2 marks)

The curve

C_2

has equation

y = -3\sin x + k

where

k

is a constant.

Given that the maximum value of

y

C_2

7

(b) (i) state the value of

k

(ii) state the coordinates of the minimum point on

C_2

with the smallest positive

x

coordinate.

(3 marks)

The curve

C_3

has equation

y = -3\sin(x + B)

where

B

is a positive constant.

The minimum point on

C_3

with the smallest positive

x

coordinate is

\left(\dfrac{\pi}{3},\ A\right)

, where

A

is a constant.
(c) Find

(i) the value of

A

(ii) the smallest possible value of

B

.

(2 marks)
(d) Using Figure 1, state the number of solutions of the equation

6\sin x + x = 0 \qquad -2\pi \leqslant x \leqslant 2\pi

giving a reason for your answer.

(2 marks)

●●●●●Level 49 marksStart

Trigonometric identities

19 questions

LessonNot started

Solve, for

0 \leqslant x \leqslant \pi

, the equation

\sin 2x = \tan x

giving your answers in terms of

\pi

. (5 marks)

●●●●●Level 45 marksStart

(a) Show that the equation

6\cos^2\theta + \sin\theta - 5 = 0

can be written as

6\sin^2\theta - \sin\theta - 1 = 0

. (2 marks)
(b) Hence solve

6\cos^2\theta + \sin\theta - 5 = 0

for

0^\circ \le \theta < 360^\circ

, giving your answers to 1 decimal place where appropriate. (5 marks)

●●●●●Level 47 marksStart

It is given that

(5\sin\theta + 4\cos\theta)^2 + (2\sin\theta - 10\cos\theta)^2 = 58

and that

\theta

is obtuse.

Find the exact value of

\sin\theta

Fully justify your answer.

(6 marks)

●●●●●Level 46 marksStart

The curve with equation

y = \sin x°

for

-360 \le x \le 360

is shown below.(a) Point

A

on the curve has coordinates

\left(a, \dfrac{\sqrt{3}}{2}\right)

(a) (i) Find the value of

a

(ii) State the value of

\sin(180° - a°)

(3 marks)
(b) Point

B

on the curve has coordinates

\left(b, -\dfrac{\sqrt{7}}{4}\right)

(b) (i) Find the exact value of

\sin(b° - 180°)

(ii) Find the exact value of

\cos b°

(5 marks)

●●●●●Level 48 marksStart

The first three terms of a geometric series are

\cos\theta \qquad\qquad 1 + \sin\theta \qquad\qquad 4\cos\theta

(a) Show that

5\sin^2\theta + 2\sin\theta - 3 = 0

(3 marks)

Given that

0 < \theta < \dfrac{\pi}{2}

,
(b) find the value of

\theta

, giving your answer to 3 decimal places.

(2 marks)
(c) Find the sum of the first

10

terms of the series.

(3 marks)

●●●●●Level 48 marksStart

In this question you must show all stages of your working.

Solutions relying entirely on calculator technology are not acceptable.
(a) Solve, for

0 < x < \pi

, the equation

5\sin(2x + 0.6) + 2 = 0

giving your answers in radians to 2 decimal places.

(4 marks)
(b) Solve, for

-180^\circ < \theta < 180^\circ

, the equation

3\tan\theta\sin\theta = 5 + 2\cos\theta

giving your answers, in degrees, to one decimal place.

(5 marks)

●●●●●Level 49 marksStart

In this question you must show all stages of your working.

Solutions relying entirely on calculator technology are not acceptable.
(a) Show that

4\cos\theta\,(2 - \tan\theta\sin\theta) = 2 + 3\cos\theta

may be written as

4\cos^2\theta + 5\cos\theta - 6 = 0

(3 marks)
(b) Hence solve, for

0 < x < 180^\circ

4\cos 3x\,(2 - \tan 3x\sin 3x) = 2 + 3\cos 3x

giving your answers to one decimal place.

(4 marks)

●●●●●Level 47 marksStart

In this question you must show all stages of your working.

Solutions relying entirely on calculator technology are not acceptable.
(a) Solve, for

0 \leqslant x < \pi

, the equation

\sqrt{3}\tan\left(3x - \dfrac{\pi}{8}\right) = 1

giving the answers in radians in the form

k\pi

, where

k

is a rational constant to be found.

(3 marks)
(b) Solve, for

0 \leqslant \theta < 360^\circ

, the equation

4\sin\theta\tan\theta = 3 - \cos\theta

giving your answers, in degrees, to one decimal place.

(5 marks)

●●●●●Level 48 marksStart

In this question you must show detailed reasoning.

Solutions relying entirely on calculator technology are not acceptable.
(a) Show that the equation

\tan\theta\,(4\sin\theta + 1) = 6\cos\theta

can be written as

10\sin^2\theta + \sin\theta - 6 = 0

(3 marks)
(b) Hence solve, for

-\pi < x < \pi

, the equation

\tan\left(x - \dfrac{\pi}{5}\right)\left(4\sin\left(x - \dfrac{\pi}{5}\right) + 1\right) = 6\cos\left(x - \dfrac{\pi}{5}\right)

giving your answers to 3 significant figures.

(4 marks)

●●●●●Level 47 marksStart

In this question you must show all stages of your working.

Solutions relying entirely on calculator technology are not acceptable.
(a) Show that

(\tan\theta + \sin\theta)(1 - \cos\theta) \equiv \sin^2\theta\tan\theta

(4 marks)
(b) Hence solve, for

0 < x < 2\pi

, the equation

(\tan x + \sin x)(1 - \cos x) = \tan x - 2\sin x\tan x

giving your answers to 3 significant figures.

(4 marks)

●●●●●Level 48 marksStart

Solving trigonometric equations

57 questions

LessonNot started

In this question you must show all stages of your working.

Solutions relying entirely on calculator technology are not acceptable.
(a) Given that

5\sin(x - 30^\circ) = 2\cos(x - 60^\circ)

show that

\tan x = \frac{7\sqrt{3}}{9}

(4 marks)
(b) Hence or otherwise solve, for

0 \leqslant \theta < 180^\circ

, the equation

5\sin(2\theta - 30^\circ) = 2\cos(2\theta - 60^\circ)

giving your answers to one decimal place.

(4 marks)

●●●●●Level 48 marksStart

(a) Express

4\sin x - 3\cos x

in the form

R\sin(x - \alpha)

, where

R > 0

and

0 < \alpha < 90^\circ

. Give

\alpha

to 2 decimal places. (3 marks)
(b) Hence solve, for

0 \le x < 360^\circ

, the equation

4\sin x - 3\cos x = 2.5

,

giving your answers to 1 decimal place. (4 marks)
(c) Write down the maximum value of

\dfrac{20}{7 + 4\sin x - 3\cos x}

,

and justify your answer. (2 marks)

●●●●●Level 49 marksStart

Solve, for

0 < x < 360^\circ

, the equation

2\cot^2 x + 5\operatorname{cosec} x + 1 = 0

,

giving your answers to 1 decimal place. You must show that one possible value of

\operatorname{cosec} x

gives no solutions, with a reason. (7 marks)

●●●●●Level 47 marksStart

Solve, for

0 \leqslant x \leqslant \pi

, the equation

\sin 2x = \tan x

giving your answers in terms of

\pi

. (5 marks)

●●●●●Level 45 marksStart

(a) Prove that

\dfrac{1 - \cos 2x}{\sin 2x} \equiv \tan x

\quad \sin 2x \ne 0

. (4 marks)
(b) Hence solve, for

0 < \theta < 360^\circ

, the equation

\dfrac{1 - \cos 2\theta}{\sin 2\theta} = 3\sin\theta

,

giving your answers to 1 decimal place. (5 marks)

●●●●●Level 49 marksStart

In this question you must show detailed reasoning.

Solve the equation

\sin\left(2x - \dfrac{\pi}{6}\right) = \cos 2x

for

0 \le x < 2\pi

, giving your answers in exact form. (5 marks)

●●●●●Level 45 marksStart

Solve, for

-180^\circ \leqslant \theta < 180^\circ

, the equation

4\cos 2\theta + 14\sin\theta = 7

giving your answers to

1

decimal place. (7 marks)

●●●●●Level 47 marksStart

Solve the equation

2\cos\left(2x - \dfrac{\pi}{3}\right) = 1

for

0 \leqslant x \leqslant \pi

, giving your answers in terms of

\pi

. (5 marks)

●●●●●Level 45 marksStart

(a) Prove that

\dfrac{\sec\theta}{\sec\theta - \cos\theta} \equiv \operatorname{cosec}^2\theta, \qquad \theta \neq \dfrac{n\pi}{2},\; n \in \mathbb{Z}

(3 marks)
(b) Hence solve, for

0 < \theta < 2\pi

, the equation

\dfrac{\sec\theta}{\sec\theta - \cos\theta} = 4

giving your answers in terms of

\pi

. (3 marks)

●●●●●Level 46 marksStart

(a) Show that

(\sin\theta + \cos\theta)^2 \equiv 1 + \sin 2\theta

. (2 marks)
(b) Hence solve the equation

(\sin\theta + \cos\theta)^2 = \frac{3}{2}

for

0^\circ \le \theta < 360^\circ

. (4 marks)
(c) Determine the greatest possible value of

\sin\theta + \cos\theta

, justifying your answer. (2 marks)

●●●●●Level 48 marksStart

(a) Show that the equation

6\cos^2\theta + \sin\theta - 5 = 0

can be written as

6\sin^2\theta - \sin\theta - 1 = 0

. (2 marks)
(b) Hence solve

6\cos^2\theta + \sin\theta - 5 = 0

for

0^\circ \le \theta < 360^\circ

, giving your answers to 1 decimal place where appropriate. (5 marks)

●●●●●Level 47 marksStart

(a) Express

2\cos\theta + 3\sin\theta

in the form

R\cos(\theta - \alpha)

, where

R > 0

and

0^\circ < \alpha < 90^\circ

, giving the exact value of

R

and the value of

\alpha

correct to 2 decimal places. (3 marks)
(b) Hence solve the equation

2\cos\theta + 3\sin\theta = 1

for

0^\circ < \theta < 360^\circ

, giving your answers to 1 decimal place. (4 marks)
(c) Determine the minimum value of

2\cos\theta + 3\sin\theta

, and the value of

\theta

in the interval

0^\circ < \theta < 360^\circ

at which it occurs. (2 marks)

●●●●●Level 49 marksStart

In this question you must show detailed reasoning.
(a) Prove that

\tan\left(\theta + \dfrac{\pi}{4}\right) + \tan\left(\theta - \dfrac{\pi}{4}\right) \equiv 2\tan 2\theta

for all values of

\theta

for which both sides are defined. (4 marks)
(b) Hence solve the equation

\tan\left(\theta + \dfrac{\pi}{4}\right) + \tan\left(\theta - \dfrac{\pi}{4}\right) = -2\sqrt3

for

0 \leqslant \theta < \pi

, giving your answers in terms of

\pi

. (4 marks)

●●●●●Level 48 marksStart

The depth of water,

d

metres, over a sandbar at the mouth of an estuary on one day is modelled by the equation

d = 4.5 + 2.3\cos\big(29(t - 3.2)\big)^\circ,

where

t

is the number of hours after midnight.
(a) (i) Find the least depth of water over the sandbar given by this model.

(ii) Find the value of

t

when the least depth first occurs.

(3 marks)
(b) A yacht can only cross the sandbar when the depth of water is at least

3.1

d = 4.5 + 0.03r + 2.3\cos\big(29(t - 3.2)\big)^\circ,

where

r

is the rainfall, in millimetres, over the previous 24 hours.
(c) Explain how this new equation could give an improved model.

(1 mark)

●●●●●Level 49 marksStart

(a) Show that the expression

3\sin 2\theta \sec\theta - \cos 2\theta \, \operatorname{cosec}\,\theta

can be written as

8\sin\theta - \operatorname{cosec}\,\theta

where

\sin\theta \ne 0

and

\cos\theta \ne 0

(4 marks)
(b) A student is attempting to solve the equation

3\sin 2\theta \sec\theta - \cos 2\theta \, \operatorname{cosec}\,\theta = 7

for

0° \le \theta \le 360°

They use the result from part (a), and write the following incorrect solution:

Step 1

\quad 8\sin\theta - \operatorname{cosec}\,\theta = 7

Step 2

\quad 8\sin^2\theta - 1 = 7\sin\theta

Step 3

\quad 8\sin^2\theta - 7\sin\theta - 1 = 0

Step 4

\quad (8\sin\theta + 1)(\sin\theta - 1) = 0

, so

\sin\theta = -\dfrac{1}{8}

\sin\theta = 1

Step 5

\quad \theta = 90°,\ 187.2°,\ 352.8°

(b) (i) Explain why the student should reject one of their values for

\sin\theta

in Step 4.

(ii) State the correct solutions to the equation

3\sin 2\theta \sec\theta - \cos 2\theta \, \operatorname{cosec}\,\theta = 7

for

0° \le \theta \le 360°

(2 marks)

●●●●●Level 46 marksStart

(a) Prove that

\dfrac{1}{1 - \sin\theta} + \dfrac{1}{1 + \sin\theta} \equiv 2\sec^2\theta

(3 marks)
(b) Hence solve, for

0^\circ \leqslant \theta < 360^\circ

, the equation

\dfrac{1}{1 - \sin\theta} + \dfrac{1}{1 + \sin\theta} = 8

(4 marks)

●●●●●Level 47 marksStart

(a) Express

5\cos\theta - 12\sin\theta

in the form

R\cos(\theta + \alpha)

, where

R > 0

and

0^{\circ} < \alpha < 90^{\circ}

. Give

\alpha

correct to

2

decimal places. (3 marks)
(b) Hence solve the equation

5\cos\theta - 12\sin\theta = 4

for

0^{\circ} \leqslant \theta \leqslant 360^{\circ}

. (3 marks)

●●●●●Level 46 marksStart

The first three terms of a geometric series are

\cos\theta \qquad\qquad 1 + \sin\theta \qquad\qquad 4\cos\theta

(a) Show that

5\sin^2\theta + 2\sin\theta - 3 = 0

(3 marks)

Given that

0 < \theta < \dfrac{\pi}{2}

,
(b) find the value of

\theta

, giving your answer to 3 decimal places.

(2 marks)
(c) Find the sum of the first

10

terms of the series.

(3 marks)

●●●●●Level 48 marksStart

(a) Prove that

\dfrac{1 + \sec\theta}{\tan\theta + \sin\theta} \equiv \operatorname{cosec}\theta

.

(4 marks)
(b) In this question you must show detailed reasoning.

Hence solve, for

0 < \theta < 2\pi

, the equation

\frac{1 + \sec\theta}{\tan\theta + \sin\theta} = 6 - 4\sin\theta.

Give your answers correct to 3 decimal places.

(5 marks)

●●●●●Level 49 marksStart

In this question you must show detailed reasoning.
(a) Show that the equation

4\sec\theta + m\sin\theta + 5\cos\theta = 0

, where

m

is a constant, can be expressed in the form

4\tan^2\theta + m\tan\theta + 9 = 0.

(3 marks)
(b) It is given that there is only one value of

\theta

, for

0^\circ < \theta < 180^\circ

, satisfying the equation

4\sec\theta + m\sin\theta + 5\cos\theta = 0

.

Given also that

m

is negative, find this value of

\theta

, correct to 3 significant figures.

(5 marks)

●●●●●Level 48 marksStart

A particle

P

moves in a straight line. At time

t

seconds, where

t \ge 0

, the velocity of

P

v

m s⁻¹, where

v = 4\cos t + 7\sin t.

(a) Express

4\cos t + 7\sin t

in the form

R\cos(t - \alpha)

, where

R > 0

and

0 < \alpha < \tfrac12\pi

. Give the value of

\alpha

correct to 4 significant figures.

(3 marks)
(b) Hence find the two smallest positive values of

t

for which

P

is moving, in either direction, with a speed of

6

m s⁻¹.

(3 marks)

●●●●●Level 46 marksStart

In this question you must show all stages of your working.

Solutions relying entirely on calculator technology are not acceptable.
(a) Prove that

\frac{\cot\theta}{\operatorname{cosec}\theta - 1} - \frac{\cot\theta}{\operatorname{cosec}\theta + 1} \equiv 2\tan\theta \qquad \theta \neq \frac{n\pi}{2},\ n \in \mathbb{Z}

(3 marks)
(b) Hence solve, for

0 < x < 180^\circ

, the equation

\frac{4\cot x}{\operatorname{cosec} x - 1} - \frac{4\cot x}{\operatorname{cosec} x + 1} = 4\sec^2 x - 1

Give each answer, in degrees, to one decimal place.

(4 marks)

●●●●●Level 47 marksStart

In this question you must show all stages of your working.

Solutions relying entirely on calculator technology are not acceptable.
(a) Show that the equation

\frac{4\sin 2\theta}{1 + \tan^2\theta} + 8\cos^3\theta\tan\theta = 6\cos^2\theta\tan\theta

may be written as

\sin 2\theta\left(A\cos^2\theta + B\cos\theta + C\right) = 0

where

A

B

and

C

are constants to be found.

(3 marks)
(b) Hence solve, for

540^\circ \leqslant x \leqslant 720^\circ

, the equation

\frac{4\sin 2x}{1 + \tan^2 x} + 8\cos^3 x\tan x = 6\cos^2 x\tan x \qquad x \in \mathbb{R},\ x \neq 630^\circ

(4 marks)

●●●●●Level 47 marksStart

(a) Write

2x^2 - 8x + 11

in the form

p(x + q)^2 + r

, where

p

q

and

r

are constants.

(3 marks)
(b) State the coordinates of the minimum point on the graph of

y = 2x^2 - 8x + 11

.

(2 marks)
(c) Hence deduce

the minimum value of $2\tan^2\theta - 8\tan\theta + 11$ ,
the smallest positive value of $\theta$ , in degrees, for which the minimum value occurs.

(3 marks)

●●●●●Level 48 marksStart

In this question you must show all stages of your working.

Solutions relying entirely on calculator technology are not acceptable.
(a) Solve, for

0 < x < \pi

, the equation

5\sin(2x + 0.6) + 2 = 0

giving your answers in radians to 2 decimal places.

(4 marks)
(b) Solve, for

-180^\circ < \theta < 180^\circ

, the equation

3\tan\theta\sin\theta = 5 + 2\cos\theta

giving your answers, in degrees, to one decimal place.

(5 marks)

●●●●●Level 49 marksStart

In this question you must show all stages of your working.

Solutions relying entirely on calculator technology are not acceptable.
(a) Show that

4\cos\theta\,(2 - \tan\theta\sin\theta) = 2 + 3\cos\theta

may be written as

4\cos^2\theta + 5\cos\theta - 6 = 0

(3 marks)
(b) Hence solve, for

0 < x < 180^\circ

4\cos 3x\,(2 - \tan 3x\sin 3x) = 2 + 3\cos 3x

giving your answers to one decimal place.

(4 marks)

●●●●●Level 47 marksStart

The first, second and fourth terms of an arithmetic progression are

u_1

u_2

and

u_4

respectively, where

u_1 = 3\sin\theta, \qquad u_2 = 5\cos\theta, \qquad u_4 = -21\sin\theta

and

\tfrac12\pi < \theta < \pi

.
(a) Determine the exact value of

\theta

.

(3 marks)
(b) Hence determine the value of

\displaystyle\sum_{r=1}^{20} u_r

.

(3 marks)

●●●●●Level 46 marksStart

A biologist models the number,

N

, of water fleas in a pond

t

days after she starts observing them by

N = 5200 - 1800\cos(15t)^\circ.

(a) State the greatest number of water fleas in the pond according to the model.

(1 mark)
(b) Write down the difference between the greatest and least values of

N

.

(1 mark)
(c) Without giving any numerical values, describe briefly how the number of water fleas changes as

t

increases.

(1 mark)
(d) Find how long it takes for

N

to first return to its value at

t = 0

.

(2 marks)
(e) Determine the time on the second occasion when

N = 6000

●●●●●Level 410 marksStart

In this question you must show all stages of your working.

Solutions relying entirely on calculator technology are not acceptable.
(a) Solve, for

0 \leqslant x < \pi

, the equation

\sqrt{3}\tan\left(3x - \dfrac{\pi}{8}\right) = 1

giving the answers in radians in the form

k\pi

, where

k

is a rational constant to be found.

(3 marks)
(b) Solve, for

0 \leqslant \theta < 360^\circ

, the equation

4\sin\theta\tan\theta = 3 - \cos\theta

giving your answers, in degrees, to one decimal place.

(5 marks)

●●●●●Level 48 marksStart

In this question you must show detailed reasoning.

Solutions relying entirely on calculator technology are not acceptable.
(a) Show that the equation

\tan\theta\,(4\sin\theta + 1) = 6\cos\theta

can be written as

10\sin^2\theta + \sin\theta - 6 = 0

(3 marks)
(b) Hence solve, for

-\pi < x < \pi

, the equation

\tan\left(x - \dfrac{\pi}{5}\right)\left(4\sin\left(x - \dfrac{\pi}{5}\right) + 1\right) = 6\cos\left(x - \dfrac{\pi}{5}\right)

giving your answers to 3 significant figures.

(4 marks)

●●●●●Level 47 marksStart

In this question you must show all stages of your working.

Solutions relying entirely on calculator technology are not acceptable.

Figure 1Figure 1 shows a sketch of the curve with equation

y = \text{f}(x)

, where

\text{f}(x) = 3\cos x\,\text{e}^{2\sin x} \qquad 0 \leqslant x \leqslant \pi

The curve intersects the

x

-axis at the point

R

, as shown in Figure 1.
(a) State the coordinates of

R

(1 mark)

The curve has two turning points, at the point

P

and the point

Q

, also shown in Figure 1.
(b) Show that, at

P

and at

Q

a\sin^2 x + b\sin x + c = 0

where

a

b

and

c

are integers to be found.

(4 marks)
(c) Hence find the

x

coordinate of

Q

, giving your answer to 3 decimal places.

(2 marks)

●●●●●Level 47 marksStart

In this question you must show all stages of your working.

Solutions relying entirely on calculator technology are not acceptable.
(a) Show that

(\tan\theta + \sin\theta)(1 - \cos\theta) \equiv \sin^2\theta\tan\theta

(4 marks)
(b) Hence solve, for

0 < x < 2\pi

, the equation

(\tan x + \sin x)(1 - \cos x) = \tan x - 2\sin x\tan x

giving your answers to 3 significant figures.

(4 marks)

●●●●●Level 48 marksStart

Radian measure

31 questions

LessonNot started

A landscape designer is planning a flower bed in the shape of a sector of a circle with radius

r

metres and angle

\theta

radians. The perimeter of the bed must be exactly

40

metres of edging.(a) Show that the area of the bed,

A

m², satisfies

A = 20r - r^2

. (3 marks)
(b) Hence find the maximum possible area of the bed, justifying that your value is a maximum, and find the value of

\theta

for which it occurs. (4 marks)

●●●●●Level 47 marksStart

Solve the equation

2\cos\left(2x - \dfrac{\pi}{3}\right) = 1

for

0 \leqslant x \leqslant \pi

, giving your answers in terms of

\pi

. (5 marks)

●●●●●Level 45 marksStart

In this question you must show all stages of your working.

Solutions relying entirely on calculator technology are not acceptable.
(a) Solve, for

0 < x < \pi

, the equation

5\sin(2x + 0.6) + 2 = 0

giving your answers in radians to 2 decimal places.

(4 marks)
(b) Solve, for

-180^\circ < \theta < 180^\circ

, the equation

3\tan\theta\sin\theta = 5 + 2\cos\theta

giving your answers, in degrees, to one decimal place.

(5 marks)

●●●●●Level 49 marksStart

Figure 1

Diagram NOT accurately drawn

The shaded area in Figure 1 shows the plan view of a café terrace.

The terrace consists of a sector

AOB

of a circle centre

O

joined to a triangle

OBC

.

Given that

$OA = OB = 5$ m
$OC = 8$ m
$BC = 6$ m
$AOC$ is a straight line

(a) show that angle

BOC

0.847

radians to 3 decimal places.

(3 marks)
(b) Find the total area of the terrace. Give your answer in m

^2

to 3 significant figures.

(3 marks)
(c) Find the perimeter of the terrace. Give your answer in metres to 3 significant figures.

(3 marks)

●●●●●Level 49 marksStart

Figure 1

Diagram NOT accurately drawn

Figure 1 shows the plan view of a school playground.

The playground consists of a sector

COD

of a circle centre

O

joined to a sector

AOB

of a different circle, also with centre

O

. The point

B

lies on the line

OC

.

Given that

angle $COD = 0.65$ radians
angle $AOB = 1.9$ radians
arc length $CD = 9.1$ m
$OB : BC = 7 : 8$

(a) show that

OB = 6.53

m to 3 significant figures.

(3 marks)
(b) Find the perimeter of the playground, in m, to 3 significant figures.

(3 marks)
(c) Find the total area of the playground, giving your answer in m

^2

to the nearest whole number.

(3 marks)

●●●●●Level 49 marksStart

In this question you must show all stages of your working.

Solutions relying entirely on calculator technology are not acceptable.
(a) Solve, for

0 \leqslant x < \pi

, the equation

\sqrt{3}\tan\left(3x - \dfrac{\pi}{8}\right) = 1

giving the answers in radians in the form

k\pi

, where

k

is a rational constant to be found.

(3 marks)
(b) Solve, for

0 \leqslant \theta < 360^\circ

, the equation

4\sin\theta\tan\theta = 3 - \cos\theta

giving your answers, in degrees, to one decimal place.

(5 marks)

●●●●●Level 48 marksStart

In this question you must show detailed reasoning.

Solutions relying entirely on calculator technology are not acceptable.
(a) Show that the equation

\tan\theta\,(4\sin\theta + 1) = 6\cos\theta

can be written as

10\sin^2\theta + \sin\theta - 6 = 0

(3 marks)
(b) Hence solve, for

-\pi < x < \pi

, the equation

\tan\left(x - \dfrac{\pi}{5}\right)\left(4\sin\left(x - \dfrac{\pi}{5}\right) + 1\right) = 6\cos\left(x - \dfrac{\pi}{5}\right)

giving your answers to 3 significant figures.

(4 marks)

●●●●●Level 47 marksStart

In this question you must show all stages of your working.

Solutions relying entirely on calculator technology are not acceptable.
(a) Show that

(\tan\theta + \sin\theta)(1 - \cos\theta) \equiv \sin^2\theta\tan\theta

(4 marks)
(b) Hence solve, for

0 < x < 2\pi

, the equation

(\tan x + \sin x)(1 - \cos x) = \tan x - 2\sin x\tan x

giving your answers to 3 significant figures.

(4 marks)

●●●●●Level 48 marksStart

Arc length and sector area

33 questions

LessonNot started

A landscape designer is planning a flower bed in the shape of a sector of a circle with radius

r

metres and angle

\theta

radians. The perimeter of the bed must be exactly

40

metres of edging.(a) Show that the area of the bed,

A

m², satisfies

A = 20r - r^2

. (3 marks)
(b) Hence find the maximum possible area of the bed, justifying that your value is a maximum, and find the value of

\theta

for which it occurs. (4 marks)

●●●●●Level 47 marksStart

In this question you must show detailed reasoning.The diagram shows a sector

AOB

of a circle with centre

O

and radius

(x + 2)

cm. The angle

AOB

is 2 radians.

The area of sector

AOB

is less than

(5x + 34)

^2

.

Find the set of possible values of

x

. Give your answer in set notation.

(5 marks)

●●●●●Level 45 marksStart

The diagram below shows a sector of a circle

OAB

.

The chord

AB

divides the sector into a triangle and a shaded segment.

Angle

AOB

\theta

radians.(a) Given that the area of the shaded segment is one third of the area of the sector

OAB

, show that

2\theta = 3\sin\theta

(4 marks)
(b) Use a suitable change of sign to show that a solution to the equation

2\theta = 3\sin\theta

lies in the interval given by

\theta \in [1.4, 1.6]

(2 marks)

●●●●●Level 46 marksStart

A circular metal disc has centre

O

and radius

9

cm.

The points

A

B

and

C

lie on the circumference of the disc so that triangle

ABC

is equilateral.

The triangle

ABC

is cut out of the disc, leaving three identical pieces of metal.
(a) Show that angle

AOB = \dfrac{2\pi}{3}

radians.

(2 marks)
(b) Find the exact total area of the three pieces of metal, giving your answer in the form

(a\pi + b\sqrt{3})

^2

, where

a

and

b

are rational numbers.

(4 marks)

●●●●●Level 46 marksStart

A metalworker is making a decorative garden screen.

The plan view of the screen is modelled as the region

OABC

, where

$OAB$ is a sector of a circle with centre $O$ and radius $r$ cm
angle $AOB = 0.75$ radians
the arc $AB$ has length $10.5$ cm
$OBC$ is a triangle joined to the sector along the radius $OB$ , where $BC = 9$ cm and angle $OBC = 2.1$ radians